Let $\operatorname{GL}_2(\mathbb{F}_5)$ be the group of invertible $2\times 2$ matrices over $\mathbb{F}_5$, and $S_n$ be the group of permutations of $n$ objects.
What is the least $n\in\mathbb{N}$ such that there is an embedding
(injective homomorphism) of $\operatorname{GL}_2(\mathbb{F}_5)$ into
$S_n$?
Such a question has been asked today during an exam; it striked me as quite difficult. There is an obvious embedding with $n=24$, and since $|\operatorname{GL}_2(\mathbb{F}_5)|=480$ and in $\operatorname{GL}_2(\mathbb{F}_5)$ there are many elements with order $20$, we have $n\geq 9$. However, "filling the gap" between $9$ and $24$ looks hard, at least to me. Can someone shed light on the topic? I would bet that representation theory and Cayley graphs may help, but I am not so much confident to state something non-trivial. I think that proving that $\operatorname{GL}_2(\mathbb{F}_5)$ is generated by three elements (is this true?) may help, too.
I would be interested also in having a proof of something sharper than $9\leq n\leq 24$.
Update. The following Wikipedia page claims, in paragraph Exceptional isomorphisms, that $\operatorname{PGL}(2,5)$ is isomorphic to $S_5$. This seems to suggest that $\operatorname{GL}_2(\mathbb{F}_5)$ embeds in $\mathbb{Z}_4\times S_5$ that embeds in $S_9$. Am I right?
Second Update. No, I am wrong, since $\mathbb{F}_{25}^*$ embeds in $\operatorname{GL}(2,\mathbb{F}_5)$, so there is an element with order $24$, so $n\geq 11$.
Best Answer
The answer is $24$. The natural action on ${\mathbb F}_5 \setminus \{0\}$ shows that ${\rm GL}_2(5) < S_{24}$.
To show that this is the smallest possible we prove the stronger result that $24$ is the smallest $n$ with $G:={\rm SL}_2(5) \le S_n$. The centre $Z = \{ \pm I_2 \}$ of $G$ has order $2$ and, since $G/Z \cong {\rm PSL}_2(5) \cong A_5$ is simple, $Z$ is the only nontrivial proper normal subgroup of $G$. So any non-faithful permutation action of $G$ has $Z$ in its kernel. It follows that the smallest degree faithful representation is transitive, and so it is equivalent to an action on the cosets of a subgroup $H < G$ with $H \cap Z = 1$. Hence we are looking for the largest subgroup $H$ of $G$ with $H \cap Z = 1$.
Since $ -I_2$ is the only element of order $2$ in $G$, all subgroups of $G$ of even order contain $Z$. There is no subgroup of order $15$, so the largest odd order subgroup has order $5$, and the permutation action on its cosets has degree $120/5 = 24$.
In general, for a finite group $G$ with a complicated structure, the problem of finding the least $n$ with $G \le S_n$ seems to be very difficult, and I have not come across any computer algorithms that solve it efficiently. The difficulty comes from the fact that the the smallest $n$ does not generally come from a transitive action, so you have to look at all possibilities of combining transitive actions to get trivial intersectino of kernels. In this particular case, we are lucky in that we can reduce the problem to ${\rm SL}_2(5)$, where we are guaranteed that the minimal action is transitive, so then it just becomes a search for the largest core-free subgroup, which can be done computationally if the group is not too huge.